Reading a book in which the author conditions on both sides of the equation of non-conditional probability and states that they still hold after conditioning without further proving. I am wondering if this is always true.
Assume we have identity:
$$\mathbb{P}(f(A_1,A_2,\ldots, A_n))=F(\mathbb{P}(f_1(A_1,A_2,\ldots, A_n)),\mathbb{P}(f_2(A_1,A_2,\ldots, A_n)),\ldots,\mathbb{P}(f_m(A_1,A_2,\ldots, A_n)))$$
where $A_i, i=1,2,\ldots,n$ are some events; $f,f_1,\ldots,f_m$ are functions on sets using set operations including $\cup$, $\cap$, $\setminus$;$F$ is a function defined on $[0,1]^m$. Then if I condition both sides on event X, do I have:
$$\mathbb{P}(f(A_1,A_2,\ldots, A_n)\mid X)=F(\mathbb{P}(f_1(A_1,A_2,\ldots, A_n)\mid X),\mathbb{P}(f_2(A_1,A_2,\ldots, A_n)\mid X),\ldots,\mathbb{P}(f_m(A_1,A_2,\ldots, A_n)\mid X))$$
Step 2: sometimes the original equation is already conditional as below:
$$\mathbb{P}(f(A_1,A_2,\ldots, A_n)\mid g(A_1,A_2,\ldots, A_n))=F(\mathbb{P}(f_1(A_1,A_2,\ldots, A_n)\mid g_1(A_1,A_2,\ldots, A_n)),\mathbb{P}(f_2(A_1,A_2,\ldots, A_n)\mid g_2(A_1,A_2,\ldots, A_n)),\ldots,\mathbb{P}(f_m(A_1,A_2,\ldots, A_n)\mid g_m(A_1,A_2,\ldots, A_n)))\tag{*}$$
then do I have the (further) conditional version:
$$\mathbb{P}(f(A_1,A_2,\ldots, A_n)\mid g(A_1,A_2,\ldots, A_n)\cap X)=F(\mathbb{P}(f_1(A_1,A_2,\ldots, A_n)\mid g_1(A_1,A_2,\ldots, A_n)\cap X),\mathbb{P}(f_2(A_1,A_2,\ldots, A_n)\mid g_2(A_1,A_2,\ldots, A_n)\cap X),\ldots,\mathbb{P}(f_m(A_1,A_2,\ldots, A_n)\mid g_m(A_1,A_2,\ldots, A_n)\cap X))$$
If any term of the original equation (*) is not conditional, simply use $f(...|X)$.
Step 3: Reverse the above statement, if "conditional version" holds true, how about the "unconditional version"?
If the above statements hold true, I can basically insert/remove conditional probability at will whenever I have an equation at hands, e.g.,
Since $\mathbb{P}(A\setminus B)=\mathbb{P}(A)-\mathbb{P}(A\cap B)$, we have $\mathbb{P}(A\setminus B\mid X)=\mathbb{P}(A\mid X)-\mathbb{P}(A\cap B\mid X)$;
Since $\mathbb{P}(A)=\mathbb{P}(B_1)\mathbb{P}(A\mid B_1)+\mathbb{P}(B_2)\mathbb{P}(A\mid B_2)$ where $B_1$ and $B_2$ is a partition of sample space, then $\mathbb{P}(A\mid X)=\mathbb{P}(B_1\mid X)\mathbb{P}(A\mid B_1 \cap X)+\mathbb{P}(B_2\mid X)\mathbb{P}(A\mid B_2 \cap X)$
Since $\mathbb{P}(A\mid B\cap C) = \frac{\mathbb{P}(A\cap B\mid C)}{\mathbb{P}(B\mid C)}$, we have $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$
There are many posts (like this and this) proving that the above statement is correct for a specific equation, but I haven't found a post handling the issue in a general way.